Algebra Word Problems, Part II: Real World Problems.
In this type of world situations, you will need to establish every variable in the situation as well as all fixed values. You generally will be given a relationship between the variable or variables.
“Richard wants to buy a shirt that is on sale for 20% off the regular price. Write the expression which represents the sale price of the shirt”.
In this situation, there are two variables: regular price and sales price. Accordingly, there is a fixed value which is a rate of change: the 20% off.
Start by writing the relationship between the variables (operation) using words:
Sales Price = Regular Price – 20% off the regular price.
The sales price is going to be the regular price minus the 20% off that regular price. Now you can substitute any symbol/variable in their place. In this case I will use s, in place of the Sales Price; and, p as the regular Price.
Substituting the variables, we get the expression:
s = r – 0.20r
Note the 0.20 is multiplied by r. This is because the 20% is off the regular price. 20% expressed as a decimal is 0.20.
“Joanna received a 7% raise at her job. She made x dollars per hour before. Write the expression that describes how much she makes now”.
The variables in this situation are two: the money she makes before and the money she makes after her raise.
Money She Makes Now = Money she made before + 7% of Money she made before
n = x + 0.07x
“The land area of Texas is about 50,000 square miles smaller than twice the land of California. Write the expression that determines the land area of Texas.”
This situation has two variables as well: land area of Texas and land area of California.
Setting up the situation, and reading the operations in inverse order as per the first lesson.
“Land area of Texas” = “50,000 smaller than twice California”
t = 2c – 50000.